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https://raw.githubusercontent.com/njcodernoob/codility-java-solutions/master/TapeEquilibrium.java
package com.challenges;

/*
 * A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.

Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

For example, consider array A such that:

  A[0] = 3
  A[1] = 1
  A[2] = 2
  A[3] = 4
  A[4] = 3
We can split this tape in four places:

P = 1, difference = |3 − 10| = 7 
P = 2, difference = |4 − 9| = 5 
P = 3, difference = |6 − 7| = 1 
P = 4, difference = |10 − 3| = 7 
Write a function:

class Solution { public int solution(int[] A); }

that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.

For example, given:

  A[0] = 3
  A[1] = 1
  A[2] = 2
  A[3] = 4
  A[4] = 3
the function should return 1, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [2..100,000];
each element of array A is an integer within the range [−1,000..1,000].
 * 
 * 
 */

public class TapeEquilibrium {
	
	public static int solution(int[] a) {
        // write your code in Java SE 8
        int sumA = 0;
        int suml = a[0];
        
        for (int i = 0; i < a.length; i++) {
            sumA += a[i];
        }
        
        int sumr = sumA - a[0];
        int minDiff = Math.abs(suml - sumr);
        
        
        for (int p = 1; p < a.length - 1; p++) {
            suml += a[p];
            sumr -= a[p];
            
            int diff = Math.abs(suml - sumr);
            
            if (diff < minDiff) minDiff = diff;
        }
        
        return minDiff;
    }

	public static void main(String[] args) {
		int[] a = {1,1};
		System.out.println(solution(a));
	}

}
